Expectation, variance etc for uniform distribution

Question

Suppose $U$ is uniformly distributed on $[0,2]$

Now I have to calculate

1. expectation
2. variance
3. distribution function
4. density

of $U^5$

They gave me a hint to use the transformation rule for expectation which is

$$\mathbb{E}[h(X)]=\int_l^rh(a)f(a)da$$ X is a random variable and $h:S\rightarrow\mathbb{R}$ and we have $[l,r]$

But how can I use this rule to calculate the expectation without having the density $f(a)da$?

My approach was to calculate the density first.

$$U\sim unif[0,2]$$ $$X:=U^5$$ $$\mathbb{P}(X\le x) = \mathbb{P}=\left(U\le \sqrt[5]{x}\right)$$ $$=\frac{1}{5}\frac{1}{\sqrt[4]{x^5}}=\int_o^xf(a)da, \ (0\le x\le32)$$ $$ f(x) = \begin{cases} -\frac{4}{25}\frac{1}{\sqrt[9]{x^5}}, & \text{$0\le x \le 32$} \\ 0, & \text{else} \end{cases}$$

Will be this the correct approach or will there be a easier way? I am stuck at this point

Answer 1

First if $supp(U)=[0,2]$, then $supp(U^5)=[0,2^5]$

1. Expectation

The hint is correct and enough to compute the expectation of $U^5$.

$$\mathbb{E}[U^5]=\int_0^2u^5p_U(u)du,$$ where $p_U(u)$ is the pdf of $U$.

Hence,

$$\begin{align} \mathbb{E}[U^5] &=\int_0^2u^5\dfrac{1}{2}du\\ &=\dfrac{16}{3} \end{align}$$

2. Variance $$\begin{align} Var(U^5) &= \mathbb{E}[U^{10}]-(\mathbb{E}[U^5])^2\\ &=\dfrac{6400}{99} \end{align}$$

3. CDF $$\begin{align} \mathbb{P}(U^5\leq u) &= \mathbb{P}(U \leq u^{1/5})\\ &=\int_0^{u^{1/5}}\dfrac{1}{2}du\\ &=\dfrac{u^{1/5}}{2} \end{align}$$

4. PDF \begin{align} p_{U^5}(u) &= \dfrac{d\mathbb{P}(U^5\leq u)}{du}\\ &=\dfrac{1}{10}u^{-4/5} \end{align}